WARNING: The below proof is wrong, as someone brought to my attention. This proof on Math/SE is correct though more complicated. Leaving the post up in case people want to try and find my mistake. :)

Wait, what?

I know what you're thinking so let's just get it out of the way.

def f(x):
  return 0 if x < 0 else 1

"There! That's a discontinuous function in python, clearly whatever big-brained math proof you have must be wrong"

Ah, how easily we write $x \lt 0$. Note that to define a function taking real numbers as input, $x$ must have infinite precision. We can think of $f(x)$ as a Turing machine which takes in an infinite bitstring of input digits and outputs another bitstring of digits. For this post, numbers are represented in fixed-point¹.

Now consider how a Turing machine "checks" $x \lt 0$ for $x = 0$. We feed in an infinite bitstring of zeros, we output $1$ if all the inputs are zero... But then either

1. The machine never halts, as it has to check all the inputs are zero before outputting anything.
2. After a certain point the machine decides the input is "close enough to zero" and outputs $1$.

In the first case $f(0)$ wouldn't be defined at zero. In the second case, we could change the digits the machine doesn't look at without changing the output (meaning a tiny tweak to the input doesn't change the output... continuity!)

Proof Idea

At some point the Turing machine processing $x$ has output $k$ digits of the output, say, after $n$ steps. In $n$ steps the machine could (at most) look at the first $n$ digits of $x$, meaning changing any digits of $x$ beyond the $n$th won't change the output.

This is continuity! Any changes to $x$ smaller than $2^{-n}$ won't change the output.

Proof

Let $x \in \mathbf{R}$ and $\epsilon > 0$. We want to find a $\delta > 0$ so that

Let $k$ be large enough that changing any digits after the $k$th in $f(x)$ doesn't change $f(x)$ by more than $\epsilon$.

Let $n$ be how many steps it takes the Turing machine before having output the first $k$ digits of the output. Now changing any digits after the $n$th in $x$ won't change the first $k$ digits of the output, since the machine is only looking at the first $n$ in order to decide the first $k$.

Now translating into $\epsilon-\delta$ language: Let $\delta$ be small enough that $|x-y|$ implies the first $n$ digits of $x,y$ are the same. This implies the first $k$ digits of the output will be the same, meaning the outputs change by less than $\epsilon$ by our definition of $k$.

Conclusion

Hopefully I've shaken your faith in infinity and the reals making sense, there's still time to convert to Finitism before judgement day! It isn't too late!